What is expand (1-ϕ) d/(dϕ)(K(ϕ)) ?

The solution to expand (1-ϕ) d/(dϕ)(K(ϕ)) is d/(dϕ)(K(ϕ))-ϕ d/(dϕ)(K(ϕ))

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expand (1-ϕ) d/(dϕ)(K(ϕ))

expand

(1 - ϕ) \frac{d}{d ϕ} (K (ϕ))

\frac{d}{d ϕ} (K (ϕ)) - ϕ \frac{d}{d ϕ} (K (ϕ))

Solution steps

(1 - ϕ) \frac{d}{d ϕ} (K (ϕ))

Apply the distributive law:

a (b - c) = ab - a c

a = \frac{d}{d ϕ} (K (ϕ)), b = 1, c = ϕ

= \frac{d}{d ϕ} (K (ϕ)) 1 - \frac{d}{d ϕ} (K (ϕ)) ϕ

= 1 \frac{d}{d ϕ} (K (ϕ)) - ϕ \frac{d}{d ϕ} (K (ϕ))

Multiply:

1 \frac{d}{d ϕ} (K (ϕ)) = \frac{d}{d ϕ} (K (ϕ))

= \frac{d}{d ϕ} (K (ϕ)) - ϕ \frac{d}{d ϕ} (K (ϕ))

(y^{2} + 7 x y + 16 x^{2}) d x + x^{2} d y

14 (1 + e^{4 + \frac{x}{105}})

\frac{1}{( x + 2 ) ( x + 1 ) ^{2}}

\frac{11}{( x - 1 ) ( x + 2 )}

\frac{6}{u ( u + 2 )}

What is expand (1-ϕ) d/(dϕ)(K(ϕ)) ?
The solution to expand (1-ϕ) d/(dϕ)(K(ϕ)) is d/(dϕ)(K(ϕ))-ϕ d/(dϕ)(K(ϕ))